AThe acceleration is due to gravity at any given point if you look at it vertically, so \(-10 m/s^2\).
b) \(sin(25) = V_y/V\), so \(V_y = V*sin(25)\). We use \(V = V_0 + a t\) and then the final speed must be 0 because it stops at the highest point. So \(0 = V_y - 10t\). Solve for \(t\) and you get \(t = 32sin(25)/10 = 16sin(25)/5\)
c) \(Y = Y_0 + V_0t + (1/2)at^2\), and then we plug the values: \(Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2\) and we already have the time from "b)", so \(Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2\); then we just rearrange it \(Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]\) and finally \(Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20\)
